Biochemistry End-of-Year 2015 — Past Paper OmpathStudy

Bchem EOY 2015 — Past Paper Questions & Answers University of Nairobi, First Year Special/Supplementary Examinations 2014/2015 — HBC 100 / UPC 102 / VBC 100: Biochemistry. This paper carries no printed answer key; model answers below are written from standard biochemistry references, not copied from a marking scheme. --- Section A: Essay Questions (Attempt any THREE) — 30 Marks Question 1 The inner mitochondrial membrane is impermeable to many molecules except to those with specific carriers or permeases. Describe the mechanism through which extra-mitochondrial NADH is delivered into the mitochondrial matrix for oxidation in the electron transport chain. Model answer: Cytosolic NADH cannot cross the inner mitochondrial membrane directly, so its reducing equivalents are shuttled in via two main systems: Malate-aspartate shuttle (liver, kidney, heart — high-yield): cytosolic oxaloacetate is reduced to malate (using cytosolic NADH), malate crosses into the matrix via a malate-α-ketoglutarate antiporter, is reoxidized to oxaloacetate (regenerating matrix NADH), and oxaloacetate is transaminated to aspartate to be shuttled back out (since OAA itself can't cross the membrane) — net yield ≈ 2.5 ATP per cytosolic NADH. Glycerol-3-phosphate shuttle (skeletal muscle, brain): cytosolic NADH reduces DHAP to glycerol-3-phosphate, which is reoxidized at the outer face of the inner membrane by a membrane-bound FAD-linked glycerol-3-phosphate dehydrogenase, passing electrons to FADH₂/ubiquinone instead of NADH — net yield ≈ 1.5 ATP per cytosolic NADH (lower yield since it enters at the ubiquinone level, bypassing Complex I). --- Question 2 Indicate the name and write a detailed reaction that is catalysed by each of the following enzymes, clearly indicating whether the reaction is reversible or irreversible: (a) G3P to DHAP converting enzyme. (b) OAA and acetyl-CoA condensing enzyme. (c) A PEP carboxylating enzyme. (d) Isocitrate cleavage enzyme. (e) An enzyme associated with favism. Model answer: (a) Triose phosphate isomerase — Glyceraldehyde-3-phosphate ⇌ Dihydroxyacetone phosphate. Reversible. (b) Citrate synthase — Oxaloacetate + Acetyl-CoA + H₂O → Citrate + CoA-SH. Irreversible (large negative ΔG, the committed first step of the TCA cycle). (c) Pyruvate carboxylase carboxylates pyruvate, not PEP directly — but the classic "PEP carboxylating enzyme" referred to here is PEP carboxykinase (PEPCK) acting in reverse, or more precisely pyruvate carboxylase : Pyruvate + CO₂ + ATP → Oxaloacetate + ADP + Pi. Irreversible. (If the intended enzyme is PEPCK itself: Oxaloacetate → PEP + CO₂ + GDP, from GTP — also irreversible in the gluconeogenic direction.) (d) Isocitrate lyase — Isocitrate → Succinate + Glyoxylate (glyoxylate cycle). Irreversible. (e) Glucose-6-phosphate dehydrogenase (G6PD) — Glucose-6-phosphate + NADP⁺ → 6-phosphogluconolactone + NADPH + H⁺. Irreversible (committed step of the PPP oxidative phase). Deficiency of this enzyme causes favism/hemolytic anemia under oxidative stress. --- Question 3 Describe the factors which affect the rate of enzyme-catalysed reactions. Model answer: Substrate concentration — rate increases with [S] until the enzyme is saturated (Vmax). Enzyme concentration — rate is generally proportional to enzyme amount, when substrate is not limiting. Temperature — rate rises with temperature up to an optimum, beyond which the enzyme denatures and activity falls sharply. pH — each enzyme has an optimum pH; deviations alter ionization of catalytic/binding residues and can denature the enzyme. Presence of inhibitors — competitive, non-competitive, or uncompetitive inhibitors reduce rate by different mechanisms. Presence of activators/cofactors — many enzymes require metal ions or coenzymes for full activity. Product concentration — accumulating product can slow the reaction (product inhibition, or by mass-action/reversibility). --- Question 4 The Henderson-Hasselbalch equation is used to calculate the pH of a buffer solution. (a) Derive the equation from a hypothetical weak acid HA. [4 marks] (b) Calculate the amount in grams of acetic acid (CH₃CO₂H) and sodium acetate needed to make one liter of a 50 mM acetate buffer with a pH of 5.0 (Ka of acetic acid = 1.9×10⁻⁵, Na=23, C=12, O=16, H=1). [6 marks] Model answer: (a) Derivation: For HA ⇌ H⁺ + A⁻, Ka = [H⁺][A⁻]/[HA]. Rearranging: [H⁺] = Ka·[HA]/[A⁻]. Taking -log of both sides: -log[H⁺] = -log Ka - log([HA]/[A⁻]), i.e. pH = pKa + log([A⁻]/[HA]) . (b) pKa = -log(1.9×10⁻⁵) ≈ 4.72. Using pH = pKa + log([A⁻]/[HA]): 5.0 = 4.72 + log([A⁻]/[HA]) → log([A⁻]/[HA]) = 0.28 → [A⁻]/[HA] ≈ 1.9. With total buffer concentration [A⁻] + [HA] = 50 mM: [HA] ≈ 50/(1+1.9) ≈ 17.2 mM, [A⁻] ≈ 32.8 mM. Mass of acetic acid (MW ≈ 60 g/mol: C₂H₄O₂ = 2×12 + 4×1 + 2×16 = 60): 0.0172 mol × 60 g/mol ≈ 1.03 g . Mass of sodium acetate (MW ≈ 82 g/mol: C₂H₃O₂Na = 2×12 + 3×1 + 2×16 + 23 = 82): 0.0328 mol × 82 g/mol ≈ 2.69 g . --- Question 5 Explain using relevant examples how eicosano